Draw the 13 C NMR spectrum of each molecule in Assessment 15.48. (c)

Welcome back everyone to another video, provide approximate chemical shifts for all equivalent carbon sets in the given molecule. We're given our structure and to identify our approximate chemical shifts, we're going to utilize the carbon NMR table. First of all, we noticed that our compound has an internal plane of symmetry, which essentially means that carbons A&E, since they are symmetrical, they will produce the same shift. Similarly, carbons B and D, they will produce the same signal. Then we will have C we will also have a pair of FNG moving on. We have H then we have I and finally JK and L those three methyl groups are bonded to the same carbonara, meaning they will all produce the same shift. Now, starting with A&E, those are the methyl groups, the H three, we expect a range between five and 45 but oxygen is an electron, the shielding group. So instead of having a maximum of 45 we're going to go outside of the range and at about 10 P PM to get a total of 55 P PM because oxygen is an election withdrawing group and the shielding rights, we expect to have a higher value. That would be about 55 parts per million for B and D. Those are the benzene carbons, right? In a range between 120 to 150. We're going to take the higher value because oxygen is electron withdrawing and the shielding and add, let's say 10 P PM to give us an accurate estimate. So that'll be about 160. So we're going to say 160 parts per million moving on to C it's really important to understand that C is the orth opposition relative to each alcoy group. And we know that ort s position gets a higher electron density because oxygen is electron withdrawing by resonance, you are going to take the lower value. We know that that would be our benzene carbon. And from that lower value, we're going to actually subtract 20 because we have two al cox election donating groups. So we will get about 100 bars for a million F and G. What we understand about F and G is that they will also have a higher electron density. So they will be more shielded because let's say F is Ortho relative to one alcoy group and power relative to another because it's not the same Ortho position, but rather Ortho and power. When we consider the two Al Coxie groups, the predicted shift will be slightly higher than the previous one. So instead of 100 we will get about 105. Now, for age, we noticed that this is the metas position, which has a lower electron density. So it's actually the shielded and we will just take the higher value for the range of benzene. We're going to say that for age, we're just going with the benzene carbon and that will be at 150. Notice that it is slightly lower than that for B and D because that carbon age is further away from oxygen, right? And that's it. And now for I, we have a qua carbon, a questionary carbon will essentially be on a higher end of our ch range. Specifically, we have to recall that if we have a questionary carbon, we will have it at about 35 P PM. So not necessarily exceeding 45. So that would be at about 35 P PM. There are no election withdrawing groups, right? So we are still in that range and we're going to say 35 parts per million notice that let's say A&E those were at 55 P PM just because there was oxygen, which is electron withdrawing, right? But in this case, we're going to consider our questionary carbon as having no electron withdrawing groups. So we definitely want to be in that range. So even if you say 45 P PM, that's perfectly fine. And finally, for JK and L, those are our methyl groups that will be more shielded than a quad carbon because we have more ch bonds specifically. And we can say that the shift will be at around 30 P PM and that would be it. We have got our shifts. Thank you for watching.

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1. A Review of General Chemistry5h 5m
2. Molecular Representations1h 14m
3. Acids and Bases2h 46m
4. Alkanes and Cycloalkanes4h 17m
5. Chirality3h 39m
6. Thermodynamics and Kinetics1h 22m
7. Substitution Reactions1h 48m
8. Elimination Reactions2h 30m
9. Alkenes and Alkynes2h 9m
10. Addition Reactions3h 19m
11. Radical Reactions1h 58m
12. Alcohols, Ethers, Epoxides and Thiols2h 42m
13. Alcohols and Carbonyl Compounds2h 17m
14. Synthetic Techniques1h 26m
15. Analytical Techniques:IR, NMR, Mass Spect7h 3m
16. Conjugated Systems6h 13m
17. Ultraviolet Spectroscopy51m
18. Aromaticity2h 34m
19. Reactions of Aromatics: EAS and Beyond5h 1m
20. Phenols55m
21. Aldehydes and Ketones: Nucleophilic Addition4h 56m
22. Carboxylic Acid Derivatives: NAS2h 51m
23. The Chemistry of Thioesters, Phophate Ester and Phosphate Anhydrides1h 10m
24. Enolate Chemistry: Reactions at the Alpha-Carbon1h 53m
25. Condensation Chemistry2h 9m
26. Amines1h 43m
27. Heterocycles2h 0m
28. Carbohydrates5h 53m
29. Amino Acids4h 20m
30. Peptides and Proteins2h 42m
31. Catalysis in Organic Reactions1h 30m
32. Lipids 2h 50m
33. The Organic Chemistry of Metabolic Pathways2h 52m
34. Nucleic Acids1h 32m
35. Transition Metals6h 14m
36. Synthetic Polymers1h 49m

15. Analytical Techniques:IR, NMR, Mass Spect

Carbon NMR

Organic Chemistry

15. Analytical Techniques:IR, NMR, Mass Spect

Carbon NMR

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Problem 49c

Textbook Question

Draw the ¹³C NMR spectrum of each molecule in Assessment 15.48.
(c)

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Identify the molecular structure of each compound in Assessment 15.48. This involves recognizing the different types of carbon environments present in the molecule.

Determine the number of unique carbon environments. Each unique environment corresponds to a distinct signal in the ¹³C NMR spectrum.

Consider the symmetry of the molecule. Symmetrical molecules may have fewer unique carbon environments due to equivalent positions.

Predict the chemical shift range for each type of carbon. For example, carbonyl carbons typically appear downfield (higher ppm), while alkyl carbons appear upfield (lower ppm).

Sketch the ¹³C NMR spectrum, placing each predicted signal at the appropriate chemical shift range. Ensure that the number of signals matches the number of unique carbon environments identified.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

¹³C NMR Spectroscopy

¹³C NMR spectroscopy is a technique used to determine the structure of organic compounds by observing the magnetic environment of carbon-13 nuclei. Each unique carbon environment in a molecule produces a distinct signal in the spectrum, allowing chemists to deduce the carbon framework of the compound. Understanding chemical shifts, signal splitting, and integration is crucial for interpreting ¹³C NMR spectra.

Chemical Shift

Chemical shift in NMR spectroscopy refers to the resonant frequency of a nucleus relative to a standard in a magnetic field. In ¹³C NMR, chemical shifts provide information about the electronic environment surrounding carbon atoms. Factors such as electronegativity of nearby atoms and hybridization state influence the chemical shift, helping to identify functional groups and structural features.

Multiplicity and Coupling

Multiplicity in NMR refers to the splitting of NMR signals into multiple peaks due to spin-spin coupling between neighboring nuclei. In ¹³C NMR, coupling with hydrogen atoms (¹H) can lead to splitting patterns, although often decoupled spectra are recorded to simplify analysis. Understanding coupling helps in determining the connectivity and spatial arrangement of atoms within a molecule.

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